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Let $RG$ denote a group ring and let $\epsilon: RG \to R $ be the augmentation map. Then $\ker(\epsilon)$, denoted $\Delta(G)$, is called augmentation ideal. If we let $\alpha= \sum_ {g \in G} a_g.g $ be an element of $\Delta(G)$ then it can be shown easily that $\alpha = \sum_{g \in G} a_g.(g-1)$ and $\Delta(G)=\{ \sum_{g\in G}a_g(g-1): g \in G, g \neq 1, a_g \in R\}$.

1) My first question is what does $(g-1) \in G$ mean? Suppose $G$ is $S_3$. Then what does $(123)-1$ mean? Is it $g.1^{-1}$ in case operation on $G$ is not additive?

2) Now if we have to check that $\Delta(G)$ is an ideal and we let $\beta =\sum_{g\in G} b_g.g \in RG$ so $\beta.\alpha $ must be in $\Delta(G)$ where $\alpha= \sum_{g \in G} a_g.g \in \Delta(G)$. But $\epsilon(\beta.\alpha)= \epsilon (\sum_{g,h \in G} b_ga_hgh)=\sum_{g,h \in G}b_ga_h$ might not be zro. Why will it be zero if $\alpha= \sum_ {g \in G} a_g=0$?

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  • $\begingroup$ If you know that $\Delta(G)=\ker\epsilon$, then it is clear that $\Delta(G)$ is an ideal! $\endgroup$ – Hagen von Eitzen Jan 17 '15 at 22:05
  • $\begingroup$ Yeah that I know. I just wanted to see how the computation part works in group rings. I got it now. Thanks anyways. $\endgroup$ – Bhaskar Vashishth Jan 17 '15 at 22:12
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  1. The kernel is an ideal of $RG$, so in particular its elements should be thought of as elements of the group ring. The sign $(g-1)$ here does not refer to an element of the group, but rather to the element $1_{R}.g + (-1_{R}).1_{G}$.

  2. Let $g \in G$. Then $\epsilon ((1.g).\alpha) = \epsilon ( \sum_{h \in G}(1.\alpha_{h}.g.h))$. Now, multiplication by a fixed $g \in G$ is a bijection on $G$, so the set $\{ g.h | h \in G\}$ is just equal to the set $\{ h \in G \} = G$. Therefore $$ \epsilon ( \sum_{h \in G}(1.\alpha_{h}.g.h)) = \sum_{h \in G} \alpha_{h} = \epsilon (\sum_{h \in G} \alpha_{h}.h) = \epsilon (\alpha) = 0.$$ Now $$\epsilon (\beta . \alpha ) = \epsilon ((\sum_{g \in G} \beta_{g}.g).\alpha) = \epsilon ( \sum_{g \in G} (\beta_{g}(g.\alpha)) = \sum_{g \in G}\beta_{g} \epsilon(g.\alpha),$$ as $\epsilon$ is a homomorphism. But each $\epsilon ( g.\alpha)$ is $0$, so the whole thing is $0$.

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  • $\begingroup$ Thanks a lot. It helped a lot to compute in group rings. Now I can move forward. $\endgroup$ – Bhaskar Vashishth Jan 17 '15 at 22:10

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